发布时间 : 星期四 文章数字电路与数字电子技术 课后答案第四章更新完毕开始阅读57abd6fbaef8941ea76e05aa
=ABC+ABC+A?BC =ABC+ABC+(A⊕B)C
=ABC+ABC+ABC+ABC 9.证明
(1) 如果ab+ab = c，则ac+ac = b，反之亦成立 (2) 如果ab+ab = 0，则 ax+by= ax+by 证:
(1) ac+ ac = a (ab+ab)+a(ab+ab)
= a (ab+ab)+ab = ab+ab = b
(2) ab+ab = 0 说明a =b或b =a
ax+by=ax+ay=ax?ay = (a+x)(a+y) = ax+ay+xy = ax+ay
= ax+by
10.写出下列各式F和它们的对偶式，反演式的最小项表达式 (1)F= ABCD+ACD+BCD (2)F= AB+AB+BC
(3)F= AB+C+BD+AD+B+C 解:
(1) F=∑m(4,11,12,15)
F=∑m (0,1,2,3,5,6,7,8,9,10,13,14) F`=∑m (15,14,13,12,10,9,8,7,6,5,2,1) (2) F=∑m (2,3,4,5,7)
F=∑m (0,1,6) F`=∑m (7,6,1)
(3) F= ∑m (1,5,6,7,8,913,14,15) F= ∑m (0,1,3,4,10,11,12) F`= ∑m (15,13,12,11,5,4,3) 11.将下列函数表示成最大项之积 (1) F= (A⊙B)(A+B)+(A⊙B)AB
(2) F= (A⊕B)+A(B⊕C) 解:
(1) F= (A⊙B)?(A+B+AB)
= (AB+AB)(A+B) = AB+AB = AB=∑m (3) =ΠM (0,1,2)
(2) F= (A⊕B)+A(BC+BC)
= AB+AB+ABC+ABC
= AB+AB+ABC = ∑m (1,2,3,4,5) =ΠM (0,6,7)
12. 用公式法化简下列各式 (1) F= A+ABC+ABC+BC+B 解:
F= A(1+BC+BC)+B(C+1) = A+B (2) F= ABC+ACD+AC 解:
F=AB+AC+CD
(3) F= (A+B)(A+B+C)(A+C)(B+C+D) 解:
F`= AB+ABC+AC+BCD
= AB+AC+BCD = AB+AC F``= F= (A+B)(A+C) (4) 解:
F=AB+AB?BC+BC
F= AB+AB+BC+BC = AB+AC+BC
a) F=AB+BC+B(AC+AC) 解:
F=BC+AC
(5) F= (x+y+z+w) (v+x) (v+y+z+w) 解:
F`= xyzw+vx+vyzw = vx+vyzw+xyzw = vx+vyzw
F``= F= (v+x) (v+y+z+w)
13.指出下列函数在什么输入组合时使F=0 (1) F=∑m (0,1,2,3,7) (2) F=∑m (7,8,9,10,11) 解:
(1) F在输入组合为4,5,6时使F= 0
(2) F在输入组合为0,1,2,3,8,10,11,13,14,15时使F= 0 14.指出下列函数在什么组合时使F=1 (1) F=ΠM (4,5,6,7,8,9,12) (2) F=ΠM (0,2,4,6) 解:
(1) F在输入组合为0,1,2,3,8,10,11,13,14,15时使F=1; (2) F 在输入组合为1,3,5,7时使F=1 15.变化如下函数成另一种标准形式 (1) F=∑m (1,3,7)
(2) F=∑m (0,2,6,11,13,14)
Fˊ = A + B 0 1 1 1 11
( a ) 0 1 1 1 10 F=AB+AC ( b ) A+C) F= (A+B) (⑤F= (AC+BC)( B+AC+AC)
= ABC+BC+AC AB (3) F=ΠM (0,3,6,7)
(4) F=ΠM (0,1,2,3,4,6,12) 解:
(1) F=ΠM (0,2,4,5,6)
(2) F=ΠM (1,3,4,5,7,8,9,10,12,15) (3) F=∑m (1,2,4,5)
(4) F=∑m (5,7,8,9,10,11,13,14,15) 16.用图解法化简下列各函数 (1) 化简题12中(1),(3),(5) (2) F=∑m (0,1,3,5,6,8,10,15) (3) F=∑m (4,5,6,8,10,13,14,15) (4) F=ΠM(5,7,13,15)
(5) F=ΠM (1,3,9,10,11,14,15)
(6) F=∑m (0,2,4,9,11,14,15,16,17,19,23,25,29,31)
(7) F=∑m (0,2,4,5,7,9,13,14,15,16,18,20,21,23,25,29,30,31) 解:
(1) 化简题12中(1),(3),(5)
① ③ AB AB CD 00 01 11 10 C 00 01 11 10 0 1 0 0 00 0 0 1 1 1 0 1 0 0 01 1 0 1 1 1 C 00 01 11 10 F=AC+BC
0 0 0 0 0
1 1 0 1 1 图P4.A16 ( 1 )
(2) F= ∑m (0,1,3,5,6,8,10,15) ( C )
AB
CD 00 01 11 10 00 1 1 F=ABC+ABD+ACD +ABD+ABCD+ABCD 01 1 1
11 1 1
1 1 10
(3) F=∑m (4,5,6,8,9,10,13,14,15) 图P4.A16 ( 2 )
AB
CD 00 01 11 10 00 01 11 1 1 1 1 1 1 F=ABC+ABC+ABD
+BCD+ ACD
(4) F=ΠM(5,7,13,15)
AB
CD 00 01 11 10 00 F= BD 01 0 0 F=B+D 0 0 11
10 (5) F=ΠM (1,3,9,10,11,14,15) 图P4.A16 ( 4 )
AB CD 00 01 11 10 F= AC+BD 00 F = (A+C)(B+D) 01 0 0
11 0 0 0 (6) F=∑m (0,2,4,9,11,14,15, 16,17,19,23,25,29,31) 0 0
10 图AB
CDP4.A16 ( 5 ) 000 001 011 010 110 111 101 100 00 1 1 1 1 1 1 01 1 1 1 1 1 1 11 10 1 1 F=ABDE+ABCE+ABCD+ABC(7) F=∑m (0,2,4,5,7,9,13,14,15,16,18,20,21,23,25,29,30,31) 图E+ABP4.A16 ( 6 ) DE+ACDE+ABCD+A
AB
CD 000 001 011 010 110 111 101 100 00 1 1 1 1 1 1 1 1 1 01 1 1 1 1 1 11 10 1 1 1 1 F= ACE+BDE+BCD+BCD+BCE 17. 将下列各函数化简成与非一与非表达式图P4.A16 ( 7 ) ,并用与非门实现 (1) F=∑m (0,1,3,4,6,7,10,11,13,14,15) (2) F=∑m (0,2,3,4,5,6,7,12,14,15)
CE
B